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# NCERT Solutions for Class 11 Maths Chapter 2 - Relations and Functions Exercise 2.1

Last updated date: 11th Aug 2024
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## NCERT Solutions for Exercise 2.1 Class 11 Maths Chapter 2 - FREE PDF Download

Exercise 2.1 Class 11 Maths focuses on the basic definitions and properties of relations and functions, laying the groundwork for more advanced topics in this chapter. Students will learn about ordered pairs, Cartesian products of sets, and the different types of relations. Ex 2.1 class 11 is designed to help students understand how to identify and represent relations between sets. Students can access the revised Class 11 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Exercise 2.1 Class 11 Maths Chapter 2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 2 Exercise 2.1 Class 11 | Vedantu
3. Formulas Used in Class 11 Chapter 2 Exercise 2.1
4. Access NCERT Solutions for Maths Class 11 Chapter 2 - Relations and Functions
4.1Exercise 2.1
5. Class 11 Maths Chapter 2: Exercises Breakdown
6. CBSE Class 11 Maths Chapter 2 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

These solutions are aligned with the updated Class 11 Maths Syllabus guidelines, ensuring students are well-prepared for exams. Class 11 Chapter 2 Maths Exercise 2.1 Questions and Answers PDF provides accurate answers to textbook questions and assists in effective exam preparation and better performance.

## Glance on NCERT Solutions Maths Chapter 2 Exercise 2.1 Class 11 | Vedantu

• NCERT Solution for Exercise 2.1 Class 11 Maths focuses on the topic of Cartesian Products of Sets.

• The Cartesian Product of Sets is a fundamental concept in set theory and is crucial for understanding relations and functions.

• Given two sets, say A and B, the Cartesian product, denoted as A×B, is the set of all possible ordered pairs where the first element is from set A and the second element is from set B.

• Ex 2.1 class 11 covers 10 fully solved questions and solutions.

## Formulas Used in Class 11 Chapter 2 Exercise 2.1

• Cartesian Product: A×B={(a,b)∣a∈A and b∈B}

Competitive Exams after 12th Science

## Access NCERT Solutions for Maths Class 11 Chapter 2 - Relations and Functions

### Exercise 2.1

1. If $\left( \dfrac{x}{3}+1,y-\dfrac{2}{3} \right)=\left( \dfrac{5}{3},\dfrac{1}{3} \right)$ , find the values of  $\text{x}$ and $y$ .

Ans: We are provided with the fact that $\left( \dfrac{x}{3}+1,y-\dfrac{2}{3} \right)=\left( \dfrac{5}{3},\dfrac{1}{3} \right)$

These are ordered pairs which are equal to each other, then the corresponding elements should also be equal to each other.

Thus, we will have, $\dfrac{x}{3}+1=\dfrac{5}{3}$

And also $y-\dfrac{2}{3}=\dfrac{1}{3}$

Now, we will try to simplify the given equations and find our needed values.

$\dfrac{x}{3}+1=\dfrac{5}{3}$

$\Rightarrow \dfrac{x}{3}=\dfrac{5}{3}-1$

Simplifying further,

$\Rightarrow \dfrac{x}{3}=\dfrac{5-3}{3}=\dfrac{2}{3}$

$\Rightarrow x=2$

So, we have the value of $x$ as $2$.

Again, for the second equation,

$y-\dfrac{2}{3}=\dfrac{1}{3}$

$\Rightarrow y=\dfrac{2}{3}+\dfrac{1}{3}$

And, after more simplification,

$\Rightarrow y=\dfrac{1+2}{3}=\dfrac{3}{3}$

$\Rightarrow y=1$

So, we have the values of $x$ and $y$ as $2$ and $1$ respectively.

2. If the set A has 3 elements and the set B={3,4,5}, then find the number of elements in $(A\times B)$.

Ans: We are provided with the fact that the set $A$ has $3$ elements and the set $B$ is given as $\{3,4,5\}$.

So, the number of elements in set $B$ is $3$.

Thus, the number of elements in $(A\times B)$ will be,

= Number of elements in $A\,\times$ Number of elements in $B$

$=3\times 3=9$

So, the number of elements in $(A\times B)$ is $9$.

3. If $G=\{7,8\}$ and $H=\{5,4,2\}$ , find $G\times H$ and $H\times G$ .

Ans: We have the sets $G=\{7,8\}$ and $H=\{5,4,2\}$ .

The Cartesian product of two non-empty sets $A$ and $B$ is defined as $A\times B=\{(a,b):a\in A\,and\,\,b\in B\}$

So, the value of $G\times H$ will be,

$G\times H=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}$

And similarly the value of $H\times G$ will be,

$H\times G=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}$

4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If $P=\{m,n\}$ and $Q=\{n,m\}$ , then $P\times Q=\{(m,n),(n,m)\}$ .

Ans: The statement is False.

We have the value as, $P=\{m,n\}$ and $Q=\{n,m\}$.

Thus, $P\times Q=\{(m,m),(m,n),(n,m),(n,n)\}$

(ii) If $A$ and $B$ are non-empty sets, then $A\times B$ is a non-empty set of ordered pairs $(x,y)$ such that $x\in A$ and $y\in B$.

Ans: The statement is True.

(iii) If $A=\{1,2\},B=\{3,4\}$ , then $A\times \{B\cap \varnothing \}=\varnothing$ .

Ans: The statement is True.

We know, $B\cap \varnothing =\varnothing$

Thus, we have, $A\times \{B\cap \varnothing \}=A\times \varnothing =\varnothing$.

5. If $A=\{-1,1\}$ , find $A\times A\times A$ .

Ans: For any non-empty set $A$ , the set $A\times A\times A$ is defined by,

$A\times A\times A=\{(p,q,r):p,q,r\in A\}$

Now, we are provided with the fact that, $A=\{-1,1\}$

Thus,

$A\times A\times A=\left\{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \right\}$

6. If $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$ . Find $A$ and $B$ .

Ans: We are provided with the fact that $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$

On the other hand, the Cartesian product of two non-empty sets $A$ and $B$ is defined as $A\times B=\{(a,b):a\in A\, and\,\,b\in B\}$

As we can see, $A$ is the set of all the first elements and $B$ is the set of all the second elements.

So, we will have, $A=\{a,b\}$ and $B=\{x,y\}$ .

7. Let $A=\{1,2\},B=\{1,2,3,4\},C=\{5,6\}$ and $D=\{5,6,7,8\}$ . Verify that

(i) $A\times (B\cap C)=(A\times B)\cap (A\times C)$

Ans: We are provided with 3 sets and we have to prove $A\times (B\cap C)=(A\times B)\cap (A\times C)$

To start with, we will have, $B\cap C=\varnothing$, as there are no elements in common between these sets.

Thus, we have, $A\times (B\cap C)=A\times \varnothing =\varnothing$

For the right-hand side, we have,

$A\times B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$

And similarly,

$A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$

Again, we can see there are no elements in common between these sets. So, we have, $(A\times B)\cap (A\times C)=\varnothing$

So, we get, L.H.S = R.H.S.

(ii) $A\times C$ is a subset of $B\times D$

Ans: Again, we are to verify, $A\times C$ is a subset of $B\times D$

So, we have, $A\times C=\{(1,5),(1,6),(2,5),(2,6)\}$

And similarly,

$B\times D=\{(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)\}$

We can easily see, every element of $A\times C$ is an element of $B\times D$. So, $A\times C$ is a subset of $B\times D$.

8. Let $A=\{1,2\}$ and $B=\{3,4\}$ . Write $A\times B$ . How many subsets will $A\times B$ have? List them.

Ans: We are provided with the fact that $A=\{1,2\}$ and $B=\{3,4\}$

Thus, we have, $A\times B=\{(1,3),(1,4),(2,3),(2,4)\}$

So, the set $A\times B$ has 4 elements.

Now, this is also known to us that, if a set $A$ has $n$ elements, then the number of subsets of $A$is ${{2}^{n}}$.

We can thus conclude that, $A\times B$will have ${{2}^{4}}=16$ subsets.

Now, noting down the subsets of $A\times B$ we get,

$\varnothing ,\{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\},\{(1,3),(1,4)\},$

$\{(1,3),(2,3)\},\{(1,3),(2,4)\},$

$\{(1,4),(2,3)\},\{(1,4),(2,4)\},\{(2,3),(2,4)\},$

$\{(1,3),(1,4),(2,3)\},\{(1,3),(1,4),(2,4)\},$

$\{(1,3),(2,3),(2,4)\},\{(1,4),(2,3),(2,4)\},$

$\{(1,3),(1,4),(2,3),(2,4)\}$

9. Let $A$ and $B$ be two sets such that $n(A)=3$ and $n(B)=2$ . If $(x,1),(y,2),(z,1)$ are in $A\times B$ , find $A$ and $B$ , where $x,y$ and $z$ are distinct elements.

Ans: We are provided with the fact that $n(A)=3$ and $n(B)=2$ ; and$(x,1),(y,2),(z,1)$ are in $A\times B$ .

We also know that $A$ is the set of all the first elements and $B$ is the set of all the second elements.

So, we can conclude, $A$ having elements $x,y,z$ and $B$ having elements $1,2$ .

Thus, we get, $n(A)=3,n(B)=2$.

So, $A=\{x,y,z\},B=\{1,2\}$ .

10. The Cartesian product $A\times A$ has 9 elements among which are found $(-1,0)$ and $(0,1)$ . Find the set $A$ and the remaining elements of $A\times A$.

Ans: We are provided with, $n(A\times A)=9$.

We also know that, if $n(A)=a,n(B)=b$ , then $n(A\times B)=ab$

As it is given that, $n(A\times A)=9$

It can be written as,

$n(A)\times n(A)=9$

$\Rightarrow n(A)=3$

And it is also given that $(-1,0),(0,1)$ are the two elements of $A\times A$ .

Again, the fact is also known that, $A\times A=\{(a, a):a\in A\}$. And also $-1,0,1$ are the elements of $A$.

Also, $n(A)=3$ , implies $A=\{-1,0,1\}$ .

So, $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),(1,1)$ are the remaining elements of $A\times A$ .

## Conclusion

Completing Exercise 2.1 of Chapter 2 in Class 11 Maths provides a solid understanding of the Cartesian Products of Sets. This exercise is essential as it lays the groundwork for more advanced topics in relations and functions. By practising the problems in class 11 ex 2.1, students learn how to form and interpret ordered pairs from two sets, which is a fundamental skill in set theory. Understanding these concepts prepares students for further studies in mathematics, enhancing their ability to analyse and work with complex relationships between sets.

## Class 11 Maths Chapter 2: Exercises Breakdown

 Exercise Number of Questions Exercise 2.2 9 Questions & Solutions Exercise 2.3 5 Questions & Solutions Miscellaneous Exercise 12 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 2 - Relations and Functions Exercise 2.1

1. What is meant by relation in class 11th math chapter 2 relation and function?

Relations are nothing but the collection of ordered beer which has one object from every set. The NCERT solution for class 11 Maths chapter 2 provides the students with a proper definition and analysis of relation as per this CBSE syllabus 2024-25 several examples present in solutions will help students in solving problems related to relationships without any difficulty.

2. How to find which relation is a function in Chapter 2 of NCERT solution for class 11th chapter relation and function?

A function that can relate every element that is available in a domain to only one element that is found in the range. It means that only one time a vertical line drawn by a student on a graph can pass through the x-axis. A relation can be found by using vertical line tests.

3. What is the importance of the NCERT solution for class 11 Maths chapter 2 relations and functions?

NCERT Solutions for Ex2.1 Class 11 relations and functions are created by highly qualified teachers and Math experts to promote clear math learning. The solution format is prominent in delivering a brief understanding of each concept. The language used in these solutions is pretty simple for enabling kids to build an in-depth knowledge of relations and functions. These resources are perfect for scoring good grades in exams and for various competitive studies.

4. What are the important formulas in NCERT solution class 11 ex 2.1?

NCERT solution class 11 Maths chapter 2 Relation and function exercise 2.1 explains the visualization of functions. The sum of the important concepts described in the solutions are range domain functions and algebra of relations; these are elaborated with the help of interactive examples to offer a brief and accurate understanding of each term and topic. This chapter does not have explicit formulas; rather, it requires kids to apply theoretical knowledge to solve practical problems.

5. What are the important topics covered in NCERT solution class 11 Maths chapter 2 relation and function exercise 2.1?

The important topics covered in class 11 Maths ch 2 ex 2.1 relation and function are the cartesian product of sets relations functions and algebra of real functions. The sub-topics included in these solutions are the addition of real functions, multiplication of real functions,  multiplication of real functions and the quotient of real functions.

6. What happens if one of the sets in the Cartesian Product is empty answer this according to Ex2.1 Class 11.

If either set A or B is empty, then the Cartesian Product A×B is also an empty set, denoted by ∅.

7. How many elements are in the Cartesian Product of two sets?

The number of elements in the Cartesian Product A×B is equal to the product of the number of elements in set A and set B. For example, if set A has 3 elements and set B has 2 elements, then A×B will have 3×2=6 elements. Practice more problems like this in class 11 maths ex 2.1.

8. What is the significance of the Cartesian Product in mathematics?

The Cartesian Product is significant because it forms the basis for defining relations and functions between sets. It is a foundational concept in set theory, which is essential for understanding more complex mathematical structures and relationships.

9.  Can the Cartesian Product be extended to more than two sets?

Yes, as we studied in class 11 maths ex 2.1, the Cartesian Product can be extended to more than two sets. For example, the Cartesian Product of three sets A, B, and C is denoted as A×B×C and consists of all ordered triples (a,b,c) where a∈A, b∈B, and c∈C.

10. Why is it important to understand the Cartesian Product concerning functions?

Understanding the Cartesian Product is important concerning functions because functions are essentially special types of relations between sets. The Cartesian Product helps in visualizing and defining these relations, which is important for studying and applying functions in mathematics. Practice more such kinds of problems in class 11 maths ch 2 ex 2.1.